题目
设z=f(x,y)是由z=f(x,y)所确定的函数，则z=f(x,y)
设
是由
所确定的函数，则
题目解答
答案
本题考察了复合函数求偏导的问题，难度较低
根据题目已知条件可知
是自变量
的函数，则对等式两端同时对
求偏导可得
对等式两端同时对
求偏导可得
则
故本题答案为2
解析
步骤 1：对等式两端同时对x求偏导
对等式$z={e}^{2x-3z}+2y$两端同时对x求偏导，得到
$\dfrac {\partial z}{\partial x}={e}^{2x-3z}\cdot (2-3\dfrac {\partial z}{\partial x})$
步骤 2：解出$\dfrac {\partial z}{\partial x}$
将步骤1中的等式变形，得到
$\dfrac {\partial z}{\partial x}=\dfrac {2{e}^{2x-3z}}{1+3{e}^{2x-3z}}$
步骤 3：对等式两端同时对y求偏导
对等式$z={e}^{2x-3z}+2y$两端同时对y求偏导，得到
$\dfrac {\partial z}{\partial y}={e}^{2x-3z}\cdot (-3\dfrac {\partial z}{\partial y})+2$
步骤 4：解出$\dfrac {\partial z}{\partial y}$
将步骤3中的等式变形，得到
$\dfrac {\partial z}{\partial y}=\dfrac {2}{1+3{e}^{2x-3z}}$
步骤 5：计算$3\dfrac {\partial z}{\partial x}+\dfrac {\partial z}{\partial y}$
将步骤2和步骤4中的结果代入$3\dfrac {\partial z}{\partial x}+\dfrac {\partial z}{\partial y}$，得到
$3\dfrac {\partial z}{\partial x}+\dfrac {\partial z}{\partial y}$
$=3\times \dfrac {2{e}^{2x-3z}}{1+3{e}^{2x-3z}}+\dfrac {2}{1+3{e}^{2x-3z}}$
$=\dfrac {2+6{e}^{2x-3z}}{1+3{e}^{2x-3z}}$
对等式$z={e}^{2x-3z}+2y$两端同时对x求偏导，得到
$\dfrac {\partial z}{\partial x}={e}^{2x-3z}\cdot (2-3\dfrac {\partial z}{\partial x})$
步骤 2：解出$\dfrac {\partial z}{\partial x}$
将步骤1中的等式变形，得到
$\dfrac {\partial z}{\partial x}=\dfrac {2{e}^{2x-3z}}{1+3{e}^{2x-3z}}$
步骤 3：对等式两端同时对y求偏导
对等式$z={e}^{2x-3z}+2y$两端同时对y求偏导，得到
$\dfrac {\partial z}{\partial y}={e}^{2x-3z}\cdot (-3\dfrac {\partial z}{\partial y})+2$
步骤 4：解出$\dfrac {\partial z}{\partial y}$
将步骤3中的等式变形，得到
$\dfrac {\partial z}{\partial y}=\dfrac {2}{1+3{e}^{2x-3z}}$
步骤 5：计算$3\dfrac {\partial z}{\partial x}+\dfrac {\partial z}{\partial y}$
将步骤2和步骤4中的结果代入$3\dfrac {\partial z}{\partial x}+\dfrac {\partial z}{\partial y}$，得到
$3\dfrac {\partial z}{\partial x}+\dfrac {\partial z}{\partial y}$
$=3\times \dfrac {2{e}^{2x-3z}}{1+3{e}^{2x-3z}}+\dfrac {2}{1+3{e}^{2x-3z}}$
$=\dfrac {2+6{e}^{2x-3z}}{1+3{e}^{2x-3z}}$